Question

What is the unit vector that is normal to the plane containing `( i +k)` and `(i+2j+2k)`?

Answer 1

`vecn = 2/3i + 1/3j -2/3k`

Explanation:

The vector we're looking for is `vec n = aveci+bvecj+cveck` where `vecn * (i+k)=0` AND `vecn * (i+2j+2k)=0`, since `vecn` is perpendicular to both of those vectors.

Using this fact, we can make a system of equations:

`vecn * (i+0j+k) = 0`

`(ai+bj+ck)(i+0j+k)=0`

`a+c = 0`

`vecn * (i+2j+2k) = 0`

`(ai+bj+ck) * (i+2j+2k) = 0`

`a+2b+2c = 0`

Now we have `a+c = 0` and `a+2b+2c = 0`, so we can say that:

`a+c = a+2b+2c`

`0 = 2b+c`

`therefore a+c = 2b+c`

`a = 2b`

`a/2 = b`

Now we know that `b = a/2` and `c = -a`. Therefore, our vector is:

`ai + a/2j-ak`

Finally, we need to make this a unit vector, meaning we need to divide each coefficient of the vector by its magnitude. The magnitude is:

`|vecn|=sqrt(a^2+(a/2)^2+(-a)^2)`

`|vecn|=sqrt(9/4a^2)`

`|vecn| = 3/2a`

So our unit vector is:

`vecn = a/(3/2a)i + (a/2)/(3/2a)j + (-a)/(3/2a)k`

`vecn = 2/3i + 1/3j -2/3k`

Final Answer